Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 4.70 ounces and a standard deviation of 0.40 ounce. Suppose that you select a sample of 25 oranges. a. What is the probability that the sample mean amount of juice will be at least 4.60 ounces? b. The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean? c. The probability is 77% that the sample mean amount of juice will be greater than what value?

Answer 1

Answer:

a) 0.10565

b) The two values = 4.61712 and

4.78288

c) 4.64088

Step-by-step explanation:

The z score formula we use to solve when given a random number of samples=

The formula for calculating a z-score is is z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ/√n is standard error

σ is the population standard deviation

n = random number of samples

a. What is the probability that the sample mean amount of juice will be at least 4.60 ounces?

z = (x-μ)/σ/√n

x = 4.60

μ = 4.70

σ = 0.40

n = 25

z = 4.60 - 4.70/0.40/√25

z = -0.10/0.40/5

z = -1.25

P-value from Z-Table:

P(x ≤ 4.60)

= P(x ≤ 4.60)

= P(z = -1.25)

= 0.10565

Therefore, the probability that the sample mean amount of juice will be at least 4.60 ounces is 0.10565

b. The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

The probability of 70% = 0.70

The z score for 0.70 = 1.036

This means, -1.036 and +1.036 gives a probability (0.70) or confidence level of 70%

We are to find x(raw score)

We solve using z score formula

For z = +1.036

z = (x-μ)/σ/√n

x = ?

μ = 4.70

σ = 0.40

n = 25

1.036 = x - 4.70/0.40/√25

1.036 = x - 4.70/0.08

Cross Multiply

1.036 × 0.08 = x - 4.70

0.08288 = x - 4.70

0.08288 + 4.70 = x

x = 4.78288

For z = -1.036

z = (x-μ)/σ/√n

x = ?

μ = 4.70

σ = 0.40

n = 25

-1.036 = x - 4.70/0.40/√25

-1.036 = x - 4.70/0.08

Cross Multiply

-1.036 × 0.08 = x - 4.70

-0.08288 = x - 4.70

-0.08288 + 4.70 = x

= 4.61712

Therefore, the two values = 4.61712 and

4.78288

c. The probability is 77% that the sample mean amount of juice will be greater than what value?

We are told to find the value that the sample mean is greater than

The z score for a P(x > z)

= P(x > 0.77)

= -0.739

Using z score formula, we find x first

z = (x-μ)/σ/√n

x = ?

μ = 4.70

σ = 0.40

n = 25

-0.739 = x - 4.70/0.40/√25

-0.739 = x - 4.70/0.08

Cross Multiply

-0.739 × 0.08 = x - 4.70

-0.05912 = x - 4.70

-0.05912 + 4.70 = x

x = 4.64088

Therefore, the value that the probability is 77% that the sample mean amount of juice will be greater than is 4.64088

Answer 2

The probability that the sample mean amount of juice will be at least 4.60 ounces is 0.10565, and the probability is 77% that the sample mean amount of juice will be greater than 4.64088.

Given :

• Mean of 4.70 ounces and a standard deviation of 0.40 ounces.
• Suppose that you select a sample of 25 oranges.

a) The formula of z-score can be used in order to determine the probability that the sample mean amount of juice will be at least 4.60 ounces.

$z = \dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}$

Now, substitute the values of mean, standard deviation, sample size, and 'x' in the above formula.

$z = \dfrac{4.6-4.7}{\dfrac{0.40}{\sqrt{25} }}$

$z = \dfrac{-0.1}{0.08}$

$z = -1.25$

Now, use the z-table in order to determine the p-value.

$P(x\leq 4.60)=P(z = -1.25)$

$P(x\leq 4.60)=0.10565$

b) The formula of z-score can be used in order to determine the raw score.

$z = \dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}$

Now, substitute the values of mean, standard deviation, sample size, and z in the above formula.

$1.036= \dfrac{x-4.7}{\dfrac{0.40}{\sqrt{25} }}$

Cross multiply in the above expression.

$\begin{aligned}\\1.036\times 0.08 &=x-4.70\\x &= 4.78288\\\end{aligned}$

Now, for z = -1.036 the value of 'x' can be calculated as:

$-1.036= \dfrac{x-4.7}{\dfrac{0.40}{\sqrt{25} }}$

Cross multiply in the above expression.

$\begin{aligned}\\-1.036\times 0.08 &=x-4.70\\x &= 4.61712\\\end{aligned}$

So, the two values symmetrically distributed around the population mean are 4.78288 and 4.61712.

c) The formula of z-score can be used in order to determine the raw score.

$z = \dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}$

Now, substitute the values of mean, standard deviation, sample size, and z in the above formula.

$-0.739= \dfrac{x-4.7}{\dfrac{0.40}{\sqrt{25} }}$

Cross multiply in the above expression.

$\begin{aligned}\\-0.739\times 0.08 &=x-4.70\\x &= 4.64088\\\end{aligned}$

So, the probability is 77% that the sample mean amount of juice will be greater than 4.64088.

For more information, refer to the link given below:

https://brainly.com/question/795909